A mass of 6 kg with initial velocity 16 m/s travels through a wind tunnel that exerts a constant force 8 N for a distance 1.6 m. It climbs a frictional incline of height 2.9 m inclined at an angle 16°, then moves along a second frictional surface of coefficient 0.1 before coming to rest.
The acceleration of gravity is 9.8 m/s^2. If the first frictional surface has a coefficient of 0.21 for a distance 1 m, how far does it slide along the second frictional region before coming to rest?

Answer:

Speed of Agnes during her journey was 0.848c

Explanation:

Given that;

Age of Agnes t₀ = 20 years

distance d = 2 × distance of star from Earth = 2 × 16 light-years

= 32 light-years

so get her speed speed; we use the following expression

Yvt₀ = d

( v / √(1 - ([tex]\frac{v}{c}[/tex])²) )² = ( 32 light-years / 20 yrs )²

v² / (1 - ( v²/c²)) = ( 32 × c  / 20 )²

v² / (1 - ( v²/c²)) = 2.56 × c²  

v² / c²-v²/c² = 2.56 × c²  

v²c²  / c² - v²  = 2.56 × c²  

v²  / c² - v²  = 2.56

v²  = 2.56 (c² - v²)

v²  = 2.56 (c² - v²)

v²  = 2.56c² - 2.56v²

v² + 2.56v² = 2.56c²

3.56v² = 2.56c²

v² = (2.56/3.56)c²

v = √((2.56/3.56)c²)

so v = 0.848c

Therefore, Speed of Agnes during her journey was 0.848c

Answer:

proton  nuetron electron

Answer:

B.  Energy increases with decreasing wavelength and increasing frequency.

Explanation:

Answer:

155.80rad/s

Explanation:

Using the equation of motion to find the angular acceleration:

[tex]\omega_f = \omega_i + \alpha t[/tex]

[tex]\omega_f[/tex] is the final angular velocity in rad/s

[tex]\omega_i[/tex]  is the initial angular velocity in rad/s

[tex]\alpha[/tex] is the angular acceleration

t is the time taken

Given the following

[tex]\omega_f = 6100rpm[/tex]

Time = 4.1secs

Convert the angular velocity to rad/s

1rpm = 0.10472rad/s

6100rpm = x

x = 6100 * 0.10472

x  = 638.792rad/s

[tex]\omega_f = 638.792rad/s\\[/tex]

Get the angular acceleration:

Recall that:

[tex]\omega_f = \omega_i + \alpha t[/tex]

638.792 = 0 + ∝(4.1)

4.1∝ = 638.792

∝ = 638.792/4.1

∝ = 155.80rad/s

Hence the angular acceleration as the blades slow down is 155.80rad/s

Answer:

So increasing the voltage increases the charge in direct proportion to the voltage. If the voltage exceeds the capacitors rated voltage, the capacitor may fail due to breakdown of the dielectric between the two plates that make up the capacitor.

Explanation:

A option.

Answer:

yes he is true I too believe weight only the mass never changes.

Answer:

the order of arrival is from highest to lowest

star other side of Andromeda> star near side of andromeda> other side of milky way   > center of the milky way> nevulosa orion> Pluto> Sum

Explanation:

The light that comes from stars and galaxies travels in a vacuum so its speed is constant and with a value of c = 3 108 m / s, so time will be directly proportional to the distance to the object

             x = c t

the order of arrival is from highest to lowest

star other side of Andromeda> star near side of andromeda> other side of milky way   > center of the milky way> nevulosa orion> Pluto> Sum

Answer:

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Explanation:

Statement is incomplete. Complete description is presented below:

A freight train has a mass of [tex]1.83\times 10^{7}\,kg[/tex]. The wheels of the locomotive push back on the tracks with a constant net force of [tex]7.50\times 10^{5}\,N[/tex], so the tracks push forward on the locomotive with a force of the same magnitude. Ignore aerodynamics and friction on the other wheels of the train. How long, in seconds, would it take to increase the speed of the train from rest to 80.0 kilometers per hour?

If locomotive have a constant net force ([tex]F[/tex]), measured in newtons, then acceleration ([tex]a[/tex]), measured in meters per square second, must be constant and can be found by the following expression:

[tex]a = \frac{F}{m}[/tex] (1)

Where [tex]m[/tex] is the mass of the freight train, measured in kilograms.

If we know that [tex]F = 7.50\times 10^{5}\,N[/tex] and [tex]m = 1.83\times 10^{7}\,kg[/tex], then the acceleration experimented by the train is:

[tex]a = \frac{7.50\times 10^{5}\,N}{1.83\times 10^{7}\,kg}[/tex]

[tex]a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}[/tex]

Now, the time taken to accelerate the freight train from rest ([tex]t[/tex]), measured in seconds, is determined by the following formula:

[tex]t = \frac{v-v_{o}}{a}[/tex] (2)

Where:

[tex]v[/tex] - Final speed of the train, measured in meters per second.

[tex]v_{o}[/tex] - Initial speed of the train, measured in meters per second.

If we know that [tex]a = 4.098\times 10^{-2}\,\frac{m}{s^{2}}[/tex], [tex]v_{o} = 0\,\frac{m}{s}[/tex] and [tex]v = 22.222\,\frac{m}{s}[/tex], the time taken by the freight train is:

[tex]t = \frac{22.222\,\frac{m}{s}-0\,\frac{m}{s} }{4.098\times 10^{-2}\,\frac{m}{s^{2}} }[/tex]

[tex]t = 542.265\,s[/tex]

The freight train would take 542.265 second to increase the speed of the train from rest to 80.0 kilometers per hour.

Answer:

2275000 lb.ft

Explanation:

Let work done on the cable be denoted by: W_ca

Let work done on the coal be denoted by: W_co

Now, dividing the cable into segments, let x represent the length from top of the mine shaft to the segment.

Meanwhile let δx be the length of the segment.

We are told the cable weighs 8 lb/ft. Thus;

Work done on one segment = 8 × δx × x = 8x•δx

Therefore, work done on cable is;

W_ca = ∫8x•δx between the boundaries of 0 and 650

Thus;

W_ca = 4x² between the boundaries of 0 and 650

W_ca = 4(650²) - 4(0²)

W_ca = 1,690,000 lb.ft

Workdone on the 900 lb of coal will be calculated as;

W_co = 900 × 650

W_co = 585000 lb.ft

Thus,

Total work done = W_ca + W_co

Total workdone = 1690000 + 585000

Total workdone = 1690000 + 585000

Total workdone = 2275000 lb.ft

Answer:

q₂ = -4.80 10⁻⁴  C  = - 0.48 mC, charge is negative

Explanation:

Let's use coulomb's law

           F = [tex]k \frac{q_1 q_2}{r^2}[/tex]

and the sum of forces, remember that charges of the same sign repel and of different sign attract

           ∑ F = F₁₃ + F₂₃           (1)

Let's start by fixing a reference system located at charge 1 with the positive direction to the right.  In the problem it indicates that the net force on charge 3 is F = - 20.0 N, the negative sign indicates that the force is towards the left

let's look for every force, the charge q₁ = 12 10-⁻³ C and q₃ = 15 10⁻³ C

           F₁₃ =[tex]k \frac{q_1 q_3}{x_{13}^2}[/tex]

           F₁₃ = 9 10⁹ 12.0 15.0 10⁻⁶ / (5-0)²

           F₁₃ = 64.8 10 3 N

This force is repulsive, that is, it is directed to the right

          F₂₃ = k \frac{q_2 q_3}{x_{23}^2}

          F₂₃ = 9 10⁹ q₂ 15.0 10⁻³ / (5-4)²

          F₂₃ = 135 10⁶ q₂  N

we substitute in equation 1

          -20.0 = 64.8 10³ + 135 10⁶ q₂

           q₂ = (-20 - 64.8 10³) / 135 10⁶

           q₂ = -4.80 10⁻⁴  C

the sign indicates that the charge is negative

Answer:

a = Δv/Δt

Explanation:

The little triangle stands for change and t=time and v=velocity

Answer:



a = [tex]\frac{v-u}{t}[/tex]

Answer:

Kepler's first law of planetary motion states that planets orbit the Sun in elliptical orbits, with the Sun located at one focus (option b)

Explanation:

Kepler's laws or laws of planetary motion are scientific laws that describe the movement of the planets around the Sun. The fundamental contribution of Kepler's laws was to show that the orbits of the planets are elliptical and not circular as was previously believed.

Kepler's laws are kinetic laws. This means that its function is to describe the planetary motion.

Kepler formulated three laws:

An ellipse is a closed curve that has two symmetrical axes, called foci or fixed points. In simpler words, an ellipse can be described as a flattened circle.

The degree of flattening of a closed curve is called eccentricity. When the eccentricity is equal to 0, the curve forms a perfect circle. On the other hand, when the eccentricity is greater than 0, the sides of the curve are flattened to form an ellipse.

Kepler's first law of planetary motion states that planets orbit the Sun in elliptical orbits, with the Sun located at one focus (option b)

Answer:

3065.4J

Explanation:

Given parameters:

Mass of water  = 335g

Initial temperature  = 24.5°C

Final temperature = 26.4°C

Unknown:

How much heat is absorbed by the sample  = ?

Solution:

To solve this problem, we use the expression below:

    Heat absorbed  = mass x specific heat of water x change in temperature

The specific heat of water  = 4.816J/g°C

So;

 Heat absorbed  = 335 x 4.816 x (26.4  - 24.5)  = 3065.4J

Answer:

q= 2.07*10¹⁶ C

m= 2.2*10⁸ kg.

Explanation:

a)

       [tex]G*\frac{m_{1} *m_{2}}{r_{12}^{2}} = k * \frac{q_{1}*q_{2}}{r_{12} ^{2}} (1)[/tex]

       [tex]q =\sqrt{\frac{G*m_{1}*m_{2}}{k} } = \sqrt{\frac{6.67e-11*4.69e27*1.23e25}{9e9} } = 2.07e16 C (2)[/tex]

b)

       [tex]n_{H} = \frac{q}{e} =\frac{2.07e16C}{1.6e-19C} = 1.29e35 (3)[/tex]

Answer:

Explanation:

Answer:

Summary: Nutrition and exercise are critical for improving body composition. Keeping your calories, fiber and protein in check is a good first step. All exercise can help with fat loss, but weight training is the best way to increase muscle mass.

Explanation:

Answer: TRUE

Explanation:

Answer:

Part-1: a)

Part-2: b)

Explanation:

Part-1

Part-2

Answer:

[tex]T_A=0.93hours[/tex]

[tex]S_A=5.23mph[/tex]

[tex]T_l=0.87hour[/tex]

Explanation:

From the question we are told that

Run speed of Ilya and Anya [tex]S_r=7.50mph[/tex]

Walk speed of Ilya and Anya [tex]S_w=4.00mph[/tex]

Total distance [tex]T_d=5.00miles[/tex]

Anya Walks 1/2d and runs 1/2d

ilya walks 1/2t and runs 1/2d

Generally the distance walked by Ilay k is mathematically given by

Since ilya walks 1/2t and runs 1/2t

Therefore

[tex]\frac{k}{4}=\frac{5-k}{7.5}[/tex]

[tex]k=1.7miles[/tex]

Therefore llay runing distance is m

[tex]m=5-1.7\\m=3.3mile[/tex]

a)Generally the Time walked by Anya [tex]T_A[/tex] is mathematically given by

Anya Walks 1/2d and runs 1/2d

Therefore

[tex]t=\frac{2.5}{4.}[/tex]

[tex]t_1=0.6hours[/tex]

[tex]t=\frac{2.5}{7.5}[/tex]

[tex]t_2=0.33hours[/tex]

[tex]T_A=0.6+0.33\\[/tex]

[tex]T_A=0.93hours[/tex]

b)Generally the average speed of Anya S_A is mathematically given as

[tex]S_A=\frac{Distance}{T_t}[/tex]

[tex]S_A=\frac{5.0}{0.93}[/tex]

[tex]S_A=5.23mph[/tex]

Generally the time taken b y llay [tex]T_l[/tex] is mathematically given by

[tex]T_l=2t[/tex]

[tex]T_l=2*(\frac{1.7}{4})[/tex]

[tex]T_l=0.87hour[/tex]

Answer:

Electroscope is a device which is used to test if an object is carrying charge or not. It consists of 2 metal leaves which are connected to a metal rod which in turn is connected to a disc.

Answer:

a. 62.44 m

b. 3.57 sec

c. 7.12 sec

Explanation:

The leaving speed is given as = 35 m/s

The launcher is oriented vertically, thus the angle is = 0 degrees

a) The maximum height reached is given as;

h= v²₀y / 2g

h=35² / 2 * 9.81

h= 62.44 m

b) Time the projectile reaches maximum height is given as;

 t=v₀y / g

 t=35/9.81

 t= 3.57 sec

c) The trip up is a mirror of the trip down

   This means the time the projectile will be in the air is ;

    t= 2 * 3.57

     t= 7.12 sec

Answer:

b. Specific heat increases as the number of atoms per molecule increases.

c. Specific heat at constant pressure is higher than at constant volume.

d. Monatomic gases behave like ideal gases.

Explanation:

Specific heat of the gas at constant pressure is usually higher than that of the volume.

i.e.

Cp - Cv = R

where R is usually the gas constant.

However, monoatomic gases are gases that exhibit the behavior of ideal gases. This is due to the attribute of the intermolecular forces which plays a negligible role. Nonetheless, the case is not always true for all temperatures and pressure.

Similarly, the increase in the number of atoms per molecule usually brings about an increase in specific heat. This effect is true as a result of an increase in the total number associated with the degree of freedom from which energy can be separated.

Thus, from above explanation:

Option b,c,d are correct while option (a) is incorrect.

Answer:

Part-1: a)

Part-2: b)

Explanation:

Part-1

Part-2

Answer:

Speed = 360m/s

Explanation:

Given the following data;

Wavelength = 18cm to meters = 18/100 = 0.18m

Frequency = 2000Hz

To find the speed;

Speed = frequency * wavelength

Speed = 2000 * 0.18

Speed = 360m/s

Therefore, the speed of the wave is 360 meters per seconds.

Answer:

b. The normal force between the molecules of the paper is overcome by the contact force of the hands.

Explanation:

The paper molecules are held together by a weak bond. When the student holds the paper on both sides with the center of the paper in between, the student applies two equal forces in the opposite direction of the paper making the paper molecules weaken and separate.

Answer:

The electrostatic forces from the contact of the hands with the paper causes the paper molecules to separate.

Here's the rest of the answers

1.) Which statement is correct about the strength of forces?

Electrostatic forces are trillions of times stronger than gravitational forces.

2.) A student tries to push a book across a table. Which forces acting on the book are a result of electrostatic forces?

normal force and friction

3.) What type of force can explain how a gecko climbs up a wall?

contact forces

5.) A heavy hammer is sitting on top of a board. Which statement correctly describes why the hammer does not go through the board?

The electrostatic forces between the molecules of the board prevent the force of gravity from breaking the board apart.

Answer:

3.56m due south

Explanation:

Given parameters:

Velocity of the car  = 32m/s

Time  = 9s

Unknown:

Displacement of the car  = ?

Solution:

The velocity of a body is its displacement per unit of time in a specific direction.

  Velocity  = [tex]\frac{displacement}{time}[/tex]  

 So;

  Velocity  = [tex]\frac{32}{9}[/tex]  = 3.56m due south

Answer:

x=0.12 m

Explanation:

Use the equation F=kx and isolate x because that's what you're solving for (x=F/k). Plug in your given values for F and k and solve for x [x=(2.3 N)/(19 N/m)] and you get x=0.12 m.

Answer: 90 m/s

Explanation:

Given

mass of racecar [tex]M=1.2\times10^3\ kg[/tex]

velocity of racecar [tex]u=8\ m/s[/tex]

mass of still honeybadger [tex]m=80\ kg[/tex]

after collision race car is traveling at a speed of [tex]v_1=2\ m/s[/tex]

conserving linear momentum

[tex]Mu+m\times0=Mv_1+ mv_2\quad[v_2=\text{velocity of honeybadger after colllision}][/tex]

[tex]1.2\times10^3\times8+0=1.2\times10^3\times2+80\times v_2[/tex]

[tex]1.2\times10^3(8-2)=80v_2\\v_2=\frac{7.2\times10^3}{80}=90\ m/s[/tex]

Answer:

0m

Explanation:

The displacement of a runner when he completes one lap round track in 100s is 0m.

Displacement is the distance covered in a specific direction. This quantity is referred to as a vector because it has both magnitude and direction.